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I'm just learning about transistor biasing for ac signal gains, and while I have many questions about setting Q points, I'll ask one which is not familiar to me. In the schematic below it's teaching how to bias a transistor with voltage dividers. I follow most of it until they say to set collector current by base current of .1ma. Fine, I understand HFE gains, but then proceeds to say the voltage divider across r3/r4 needs to have current approx .9ma plus the .1ma of base current. If there's no base resistor setting .1 from voltage divider, wouldn't base current be .9? If they are referring to the ac signal from coupling capacitor, how do ac voltages and currents combine with DC voltage divider bias? I can't seem to find visual descriptions to explain this interaction in principle. Any help would be appreciated.

http://www.rason.org/Projects/bipolamp/bipolamp.htm

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As far as I understand your problem, the key point is in the sentence (quote): "In order to provide a stiff base voltage, resistor R4 should have a current of about 5 to 10 times the base current. For this example, we will assume 9 times the base current for a total of .9 ma."

In general, the voltage divider is chosen with the aim to produce a base voltage Vb "as stiff as possible". That means: As independent as possible on the base current which is connected with large tolerances (due to B variations). (Due to some other constraints like power consumption and input resistance the voltage divider must not be much lower in resistance. Without these constraints, it would be even better to chose a factor between both currents of 20 or 50 instead of only 9).

This is the typical design approach because - at the same time - the emitter resistor R2 provides current controlled voltage feedback (voltage Ve) which can work satisfactorily only for a "stiff" base voltage Vb. As a consequence of the negative feedback, the collector current does not depend too much from the base-emitter voltage, which normally is chosen (for calculating purposes) to be app. Vbe=0.65 volts.

It is very easy to visualize this effect using the method of "stabilizing line" in the Ic=f(Vbe) characteristic. This is nothing else than to plot the ohmic law for the resistor R2 - expressed not by the emitter voltage Ve but by Ve=Vb-Vbe. In this case, both graphs can be plotted in one single diagram - and it becomes clear that Ic does depend only very little on the temperature-dependent Ic=f(Vbe) characteristic. The actual DC operating point exist where both curves cross each other - the exponential BJT characteristic and the linear R2 characteristic Ic=f(Vb-Vbe), which has a negative slope.

The enclosed drawing shows the construction of the stabilization line.

enter image description here

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  • \$\begingroup\$ I'll have to humbly admit that while I understood some of that, most was beyond my hobbyist knowledge at this point. Im learning transistors, not experienced with them. I do appreciate the time and effort into the explanation and I'll do my best to figure that out. You probably did answer the question, but I'm still at a loss at understanding where .1ma is generated for the base. If you care to move on, no worries and thanks again. \$\endgroup\$
    – Archaeus
    Commented Oct 7, 2016 at 15:23
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    \$\begingroup\$ Archaeus - it is NOT your goal to create a base current of 0.1mA. Instead, you want to allow a certain collector current Ic. For this purpose, you establish two voltages: Ve (using a proper Re value) and Vb (using a suitable voltage divider). The difference between both voltages must be Vbe=Vb-Ve=0.65...0.7 volts (approximately). Because of the feedback effect. the exact value is relatively unimportant. Finally, the value of Ib=Ic/B is, of course, considered during the voltage divider design (through R3 into the base node). This current Ib does exist, but it does NOT determine Ic. \$\endgroup\$
    – LvW
    Commented Oct 7, 2016 at 15:36
  • \$\begingroup\$ I understood that much better. Now I have a mental model of what the components are doing. Thank you. \$\endgroup\$
    – Archaeus
    Commented Oct 7, 2016 at 15:43
  • \$\begingroup\$ For a better understanding I have added a drawing of the stabilization effect. \$\endgroup\$
    – LvW
    Commented Oct 7, 2016 at 15:43

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